HDU 5340--three palindromes —————— "Manacher treatment palindrome string"

Three palindromes

Time limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 1244 Accepted Submission (s): 415

Problem Descriptioncan We divided a given string S into three nonempty palindromes?

Inputfirst line contains a single integerT≤ which denotes the number of test cases.

For each test case, the there is a single line contains a string S which only consist of lowercase Chinese letters.1≤| S| ≤20000

Outputfor each case with the output the "Yes" or "No" in a.

Sample Input2abcabaadada

Sample outputyesno    topic: Ask if you can find the three-segment palindrome string.  : No violent pressure, time approaching timeout. But it was a fluke.  

#include <bits/stdc++.h>using namespace std; #define MIN (a) < (b) ( A):(B)) const int Maxn=20200;int pre[maxn*2],suf[maxn*2];int P[maxn*2];char str[maxn],trans[maxn*2];int Transform () {/    /memset (P,0,sizeof (p));    memset (pre,0,sizeof (pre));    memset (suf,0,sizeof (SUF));    int Len=strlen (str);    Trans[0]= ' $ ';        for (int i=1;i<=2*len;i+=2) {trans[i]= ' # ';    TRANS[I+1]=STR[I/2];    } trans[2*len+1]= ' # ';    trans[2*len+2]= ' @ '; return 2*len+1;}    int Manacher () {int len=transform ();    int mx=0,pos=0;        for (int i=1;i<=len;i++) {if (i<mx) {p[i]=min (p[2*pos-i],mx-i);        }else{p[i]=1;        } for (; Trans[i+p[i]]==trans[i-p[i]];p [i]++);            if (Mx<i+p[i]) {mx=i+p[i];        Pos=i; }} return len;    int main () {int t;    scanf ("%d", &t);        while (t--) {scanf ("%s", str);        int Lens=strlen (str);        if (lens<3) {printf ("no\n"); continue; }else if (lens==3) {printf ("yes\n"); continue;            }else{int len= manacher ();                for (int i=2;i<len;i++) {if (p[i]==i) {pre[i+p[i]-1]=1;                } if (p[i]==len-i+1) {suf[i-p[i]+1]=1;            }} int flag=0;                    for (int i=2;i<len&& (!flag), i++) {for (int j=1;j<p[i]&& (!flag); j + +) {                        if (Pre[i-j]&suf[i+j]) {flag=1;                    printf ("yes\n");            }}} if (!flag) {printf ("no\n"); }}} return 0;}

　　

HDU 5340--three palindromes —————— "Manacher treatment palindrome string"